Is V Perpendicular a Subset of U Perpendicular in Subspaces of Rn?

[stunner5000pt]

Suppose U, V are proper subsets of Rn and are subspaces and U is a proper subset of V. PRove that V perp is a proper subset of U perp

Ok SO let U ={u1, u2, ..., un}
let V = {v1,...vn}
let V perp = {x1,x2,..., xn}
let U perp = {w1,...wn}
certainly u1 . w1 = 0
(u1 + u2 ) . (w1+w2) = 0
cu1 . cw1 = 0

everything till now is pretty much basically understood
i can't find a wayu to maek a proof work... Any hints would be appreciated!
I really want to be able to get this one!

 

[stunner5000pt]

can anyone help?

i am still stuck on this problem and i need to solve it ...

ok so far i get this part
if U perp is orthognal to U then U perp is orthogonal to V.
If V perp is orthognal to V then V perp is orthogonal to U
this V perp is orthogonal to U perp

then Wn . Xn = 0
now $$U^{\bot} \bigcap U = \{0\}$$
and $$U^{\bot} \bigcap V = \{0\}$$
similarly it applies for V perp
so both U perp and V perp include elements that are perpendicular to U and V and contain everything (but zero) that is perpendicular to U and V, thus not included in U and V.
Now how would i relate U perp and V perp to each other. Is there a theorem that says that the perpendicular space to a bigger subspace is smaller than the perpendicular subspace to the subspace contained in the bigger subspace??

 

[quicknote]

First, we need to show that V⊥ is a subset of U⊥. This means we need to prove that if x∈V⊥, then x∈U⊥.

Let x∈V⊥. This means that x⋅v=0 for all v∈V. Since U is a proper subset of V, this means that there exists at least one vector u∈U that is not in V. Therefore, x⋅u=0 since u is not in V. This shows that x∈U⊥, and thus V⊥ is a subset of U⊥.

Next, we need to show that V⊥ is a proper subset of U⊥. This means we need to prove that there exists at least one vector in U⊥ that is not in V⊥.

Since U is a proper subset of V, there exists at least one vector v∈V that is not in U. This means that v⋅u=0 for all u∈U. However, since v is not in U, there exists at least one vector u∈U such that u⋅v≠0. This shows that v∉U⊥. Therefore, there exists at least one vector in U⊥ (namely, u) that is not in V⊥. This proves that V⊥ is a proper subset of U⊥.